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Chapter 3: The “Language of Chemistry” “alphabet” symbols for the elements, e.g. C, N, F, Mg, Fe, etc. (Know names/symbols for #1-88 except lanthanides, including spelling!) “words” chemical formulas, e.g. H2O, N2, Fe2(CO3)3, etc. counting atoms in formulas: --1 molecule of H2O contains 2 hydrogen atoms and 1 oxygen --the formula Fe2(CO3)3 represents: 2 iron atoms, 3 carbon atoms, and 9 oxygen atoms “sentences” chemical equations (reactants and products) Mg(OH)2(aq) + 2 HCl(aq) MgCl2(aq) + 2 H2O(l) Coefficients are used to “balance” the equation Subscripts indicate states of matter (not always included) Balanced equation: same number of atoms of each element on both sides of arrow Molecular and Ionic Compounds • Molecular Compounds – Atoms linked together by “covalent chemical bonds” in discrete electrically neutral particles called molecules – e.g. H2O CO2 PCl3 C12H22O11 • Ionic Compounds – Result from transfer of one or more electrons from one atom to another to yield oppositely-charged particles called ions – No discrete molecules; ions held together by electrostatic forces (“ionic bonds”) in a regular, 3-D pattern called a crystalline lattice e– e.g. LiF lithium fluoride Li – MgCl2 magnesium chloride Mg + F Li+ + F- LiF e- Cl e- Cl Mg2+ + 2 ClMgCl2 Example Compounds Molecular CH4 = Ionic = = e- NaCl Na + Cl Na+ Cl- Formula Unit: The smallest unit of a compound. Shows the smallest whole-number ratio. (NaCl) Ionic compounds must be electrically neutral. Types of Chemical Formulas • empirical formula shows the simplest ratio of the elements present • molecular formula shows the actual number of atoms in one molecule • structural formula shows how the atoms are connected e.g. for “hydrogen peroxide” the three formulas are: empirical: HO molecular: H2O2 structural: H O O H Ionic Compounds • Usually involve metals and/or polyatomic ions 1- anions 2- anions 1+ cations 2+ cations Other metals may form more than one cation, e.g. Fe2+, Fe3+, Sn2+, Sn4+ Writing Formulas for Ionic Compounds • Polyatomic ions--Table 3.5--KNOW formulas and names!!! – – – – – – Two or more atoms combined in a single charged unit e.g. NH4+ (ammonium), H3O+ (hydronium), NO3- (nitrate), PO43- (phosphate), HCO3- (hydrogen carbonate, or bicarbonate) • Look for the simplest combination of cations (+) and anions (-) to yield an electrically neutral formula – e.g. – – – ion combination Mg2+ and ClNa+ and O2Fe3+ and SO42- compound MgCl2 Na2O Fe2(SO4)3 • Example: What compound should form between sulfur (S) and potassium (K)? K2S • Example: What compound will form between ammonium and phosphate? (NH ) PO 4 3 4 Nomenclature for Ionic Compounds • First, determine if it’s ionic! – metal(s) + nonmetal(s) • Binary ionic compounds (2 different elements) – cation(charge if needed) + anionide – Know Tables 3.3 and 3.4 (not older names) – e.g. ion combination compound name – Mg2+ and ClMgCl2 magnesium chloride sodium oxide – Na+ and O2Na2O iron(II) nitride – Fe2+ and N3Fe3N2 – Other ionic compounds – With polyatomic ions; cation(charge if needed) + polyatomic ion name – Hydrates; compound name (as above) + prefixhydrate (Know prefixes, p92) – e.g. ion combination compound name calcium dichromate – Ca2+ and Cr2O72CaCr2O7 – – Co2+ and ClHg22+ and CN- CoCl2•6H2O Hg2CN2•H2O cobalt chloride hexahydrate mercury(I) cyanide hydrate Nomenclature: Molecular Compounds • First, determine it’s molecular! – between nonmetals and/or metalloids • Binary molecular compounds (between 2 elements) – prefixelement + prefixelementide – First element is most metallic (bottom left of per. table) • Use prefixes to indicate numbers of each atom, e.g. – PF3 phosphorus trifluoride – P2F4 diphosphorus tetrafluoride – N2O5 dinitrogen pentoxide • Exception: hydrogen plus one atom of a nonmetal, see next section! Nomenclature; Binary Acids • First, determine it’s a binary acid! – hydrogen + nonmetal – Hydroelementic + acid – e.g. compound name hydrochloric acid – HCl – HBr hydrobromic acid Nomenclature: Oxoacids and Their Salts • oxoacid HxEOy (E = nonmetal) • Removal of H+ yields polyatomic anions oxoacid polyatomic ions salt example H2SO4 sulfuric acid SO42– sulfate Na2SO4 sodium sulfate HSO4– Hydrogen sulfate NaHSO4 Sodium hydrogen sulfate SO32– sulfite CaSO3 Calcium sulfite HSO3– Hydrogen sulfite Ca(HSO3)2 Calcium hydrogen sulfite polyprotic acids H2SO3 Sulfurous acid acid salts Series of chlorine oxoacids and their salts: HClOx (x = 1,2,3,4) Hypochlorous acid, chlorous acid, chloric acid, perchloric acid Stoichiometric Equivalence A chemical formula shows the ratio by atoms and by moles of the elements in the formula. e.g. in the compound N2O5: ratio by atoms: 2 atoms N : 5 atoms O ratio by moles: 2 moles N : 5 moles O in N2O5, 2 moles N 5 moles O (a chemical equivalence) Problem: How many moles of N atoms are combined with 15 moles of O in N2O5? Use the mole ratio as a conversion factor! (15 moles O) x (2 moles N/5 moles O) = 6.0 moles N Formula Mass and Molecular Mass • formula mass = sum of all atomic masses of elements in a formula (remember that atomic mass = the mass of a single atom) • molecular mass = formula mass of a molecular substance {“formula weight” and “molecular weight” are often used instead} Problem What is the molecular mass of N2O5? (add the atomic masses!) N2O5 = 2 N + 5 O = 2(14.0) + 5(16.0) = 108.0 What are the units? For 1 molecule: amu For 1 mole: grams 1 mole of a substance = its formula mass in grams e.g. 1 mole of N2O5 = 108.0 g N2O5 (just another conversion factor!) Example Problems • What is the mass of 0.65 moles of N2O5? (0.65 moles N2O5) x (108 g N2O5/1 mol N2O5) = 70 g N2O5 • What mass of iron combines with 5.00 g of oxygen to make Fe2O3? Method: grams A --> moles A --> moles B --> grams B (5.00 g O) x(1 mol O)/16.0 g O) x (2 mole Fe/3 mol O) x (55.85 g Fe/mole Fe) = 11.6 g Fe Sample Problems • The label on my water bottle says there are 5.0 mg of sodium in it; if that were pure sodium, how many atoms of sodium would that be? • The same water bottle contains 16 oz of water. If I drink it all, how many moles of water did I drink? Sample Problems • As I write this lecture I’m reading the label on my water bottle. It says there are 5.0 mg of sodium in it; if that were pure sodium, how many atoms of sodium would that be? • Answer: 1.3 x 1020 atoms Na • The same water bottle contains 16 oz of water. If I drink it all, how many moles of water did I drink? • Answer: 25 mol water Percentage Composition • percentage composition -- mass % of elements in a compound Theoretical % composition -- from given formula Example Problem What is the percentage composition of H2CO3? mole ratio = 2 mol H : 1 mole C : 3 mol O molecular mass = 2(1.01) + 1(12.01) + 3(15.99) = 62.00 g/mol % composition: % H = [mass H / mass H2CO3] x 100% = [2(1.01)/62.00] x 100% = 3.36% %C= = 19.36% (12.01/62.00) x 100% % O = [3 (16.00)/62.00] x 100% Total = 77.38% 100.00% Empirical Formula Determination Example Problem A certain fluorocarbon is found to be 36.52% C, 6.08% H, and 57.38% F. What is the empirical formula for this compound? We’re looking for the mole ratio of the elements. In 100 g of the compound, there are: (36.52 g C) x (1 mol C/12.01 g C) (6.08 g H) x (1 mol H/1.008 g H) (57.38 g F) x (1 mol F/19.00 g F) So, the mole ratio is: C3.041H6.02F3.020 = 3.041 mol C = 6.02 mol H = 3.020 mol F Now reduce to the simplest ratio (divide by the smallest number): C3.041/3.020H6.02/3.020F3.020/3.020 = C1.007H1.99F = CH2F (the empirical formula) Molecular Formula Empirical formula combined with molecular mass = molecular formula Problem The above fluorocarbon is found to have a molecular mass of 66.08 g/mole. What is the molecular formula? n x (mass of empirical formula) = molecular mass (n = ?) Empirical formula = CH2F Formula mass = 1 C + 2 H + F = 33.03 g/mole n x (33.03 g/mole) = 66.08 g/mol molecular formula is C2H4F2 so, n = 2 Sample Problem Carboranes are an interesting class of compounds that contain carbon, hydrogen, and boron. One such carborane is found to have the following percentage composition: 28.18% C, 63.45% B, and 8.26% H. Determine the empirical formula of this carborane. Sample Problem Carboranes are an interesting class of compounds that contain carbon, hydrogen, and boron. One such carborane is found to have the following percentage composition: 28.18% C, 63.45% B, and 8.26% H. Determine the empirical formula of this carborane. • Answer: C2B5H7 Balancing Chemical Equations I • Adjust coefficients to get equal numbers of each kind of element of both sides of arrow. • Use smallest, whole number coefficients. e.g. start with unbalanced equation (for the combustion of butane): C4H10 + O2 CO2 + H2O Hint -- first look for an element that appears only once on each side; e.g. C C4H10 + 13/2 O2 4 CO2 + 5 H2O Multiply through by 2 to remove fractional coefficient: 2 C4H10 + 13 O2 8 CO2 + 10 H2O Nomenclature: Organic Compounds • Compounds of carbon--organic chemistry – If they have only C and H, hydrocarbons – e.g. alkanes: methane CH4, ethane C2H6, propane C3H8 • general formula: CnH2n+2 – Know Table 3.7 • Functional Groups – – – – R = hydrocarbon group e.g. alcohols: methanol CH3OH, ethanol C2H5OH e.g. amines: propyl amine, butylamine Recognize functional groups in Table 3.8 Methanol (wood alcohol), CH3OH, is related to methane, CH4, by replacing one H with OH.